Question

An ideal solution consisting of 79 wt% benzene (C6H6) and 21 wt% toluene (C7H8) was heated in a closed vessel to 50 ºC. What is the mole fraction of benzene in the vapour phase when equilibrium was reached at 50 ºC? Round your answer to two significant figures.
Data: Vapour pressure of benzene at 50 ºC = 271 mmHg

Vapour pressure of toluene at 50 ºC = 91.5 mmHg

Answer 1

In this case, according to the Raoult's law, we can find the mole fraction of 79 wt% benzene and 21 wt% toluene at 50 °C in the vapor phase as follows:

$y_iP=x_iP_i$

However, we first have to calculate the mole fractions in the solution as follows (b stands for benzene and t for toluene):

$\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{(78.11*x_b+92.14*x_t)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene} \\\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{78.11*x_b+92.14*(1-x_b)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene}\\\\x_b=0.79*(x_b+1.18*(1-x_b))\\\\x_b=0.79x_b+0.932-0.932x_b\\\\x_b=\frac{0.932}{1+0.932-0.79} =0.816\\\\x_t=1-x_b=1-0.816=0.184$

Next, we calculate the total pressure as follows, according to the Dalton's law:

$P=x_bP_b+x_tP_t=0.816*271mmHg+0.184*91.5mmHg=237.972mmHg$

Finally, the mole fractions of the vapor phase turn out:

$y_b=\frac{0.816*271mmHg}{237.972mmHg}=0.929\\\\y_t=1- 0.929=0.071$

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